Question: A bus can hold a maximum of 38 students. What is the minimum number of buses needed to transport 411 students?
Explanation: Since $n$ buses will carry $38n$ students, we need $38n>411$.

Dividing both sides of this inequality by $38$, we have $n>\dfrac{411}{38}$. We can convert $\dfrac{411}{38}$ to a mixed number: $$\frac{411}{38} = \frac{380}{38}+\frac{31}{38} = 10\frac{31}{38}.$$ Since the number of buses must be an integer, the smallest possible number of buses is $\boxed{11}$.